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Sentential Calculus Revisited: Boolean Algebra

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Sentential Calculus Revisited: Boolean Algebra Sentential Calculus Revisited: Boolean Algebra Modern logic began with the work of George Boole, who treated logic algebraically. Writing “+,” “⋅,” and “-” in place of “or,” “and,” and “not,” he developed the sentential calculus in direct analogy to the familiar theories of groups, rings, and fields. Now that we have the predicate calculus with identity, we can discern the consequences of the axioms Boole wrote down, and we’ll find that Boole’s algebraic formalism got it exactly right. Here are the principles he discovered: Definition. The language of Boolean algebra is the language whose individual constants are “0” and “1,” whose function signs are the unary function sign “-” and the binary function signs “+” and “⋅,” and whose predicates are the binary predicates “=” and “≤.” For readability, we write “(x+y)” and (x⋅y)” in place of “+(x,y)” and “⋅(x,y).” So the terms of the language of Boolean algebra will constitute the smallest class of expressions which contains the variables and “0” and “1,” contains -τ whenever it contains τ , and contains (τ+ρ) whenever it contains τ and ρ. We have unique readability, as usual. Definition. A structure is a Boolean algebra just in case it satisfies the following axioms: Associative laws: (∀x)(∀y)(∀z)((x+y)+z) = (x+(y+z)) (∀x)(∀y)(∀z)((x⋅y)⋅z) = (x⋅(y⋅z)) Commutative laws: (∀x)(∀y)(x+y) = (y+x) (∀x)(∀y)(x⋅y) = (y⋅x) Idempotence: (∀x)(x+x) = x (∀x)(x⋅x) = x Distributive laws: (∀x)(∀y)(∀z)(x + (y⋅z)) = ((x+y)⋅(x+z)) (∀x)(∀y)(∀z)(x⋅(y+z)) = ((x⋅y) + (x⋅z)) Identity elements: (∀x)(x+0) = x (∀x)(x⋅1) = x Complementation laws: (∀x)(x+-x) = 1 (∀x)(x⋅-x) = 0 Boolean algebra, p. 2 Non-triviality: ¬0=1 Definition of “≤”: (∀x)(∀y)(x ≤ y ↔ (x+y)=y Definition. A sentence that’s true in every Boolean algebra is a theorem of Boolean algebra. Among the theorems of Boolean algebra are the universal closures of the following formulas: Alternative definition of “≤”: x ≤ y ↔ (x⋅y) = x Further principles governing 1 and 0: (x + 1) = 1 (x⋅0) = 0 ((∀x)(x+y) = y ↔ y=1) ((∀x)(x⋅y = y ↔ y=0) ((∀x)(x+y) = x ↔ y=0) ((∀x)(x⋅y = x ↔ y=1) Further principles governing “-”: -1 = 0 -0 = 1 --x = x (x = -y ↔ ((x+y) = 1 ∧ (x⋅y) = 0)) x ≤ y ↔ -y ≤ -x De Morgan’s laws: -(x+y) = (-x⋅-y) -(x⋅y) = (-x + -y) Absorption principles: (x + (x⋅y)) = x (x⋅(x+y)) = x ≤ is reflexive: x ≤ x ≤ is transitive: ((x ≤ y ∧ y ≤ z) → x ≤ z) Boolean algebra, p. 3 ≤ is antisymmetric: ((x ≤ y ∧ y ≤ x) → x = y) Lattice principles: x ≤ 1 0 ≤ x ((∀x)x ≤ y ↔ y = 1) ((∀x)y ≤ x ↔ y = 0) (x ≤ (x+y) ∧ y ≤ (x+y)) ((x⋅y) ≤ x ∧ (x⋅y) ≤ y) ((x ≤ z ∧ y ≤ z) ↔ (x+y) ≤ z) ((z ≤ x ∧ z ≤ y) ↔ z ≤ (x⋅y)) Further characterizations of ≤: x ≤ y ↔ (-x + y) = 1 x ≤ y ↔ (x ⋅ -y) = 0 (x ≤ y ↔ (∃z)(y⋅z)=x) (x ≤ y ↔ (∃z)(∃w)(x+z) = (y⋅w) Tests for equality: (x = y ↔ ((x + -y) = 1 ∧ (-x + y) = 1)) (x = y ↔ ((x ⋅ -y) =0 ∧ (-x ⋅ y) = 0)) (x = y ↔ (x+y) = (x⋅y)) The simplest example is the two-element Boolean algebra, whose elements are the integers 0 and 1. “0” denotes 0, and “1” denotes 1. The sum of two elements of the domain is defined to be their maximum, the product is their minimum, and the complement of x is defined to be 1 - x. In a table, a b (a+b) (a⋅b) -a 1 1 1 1 0 1 0 1 0 0 0 1 1 0 1 0 0 0 0 1 Definition. The augmented language of Boolean algebra is the language obtained from the language of Boolean algebra by adding infinitely many additional constants c0, c1, c2, c3,.... Definition. For τ a term of the augmented language of Boolean algebra, the dual of τ, τd, is the term obtained from τ by exchanging “0” and “1” everywhere and exchanging “+” and “⋅” everywhere. Boolean algebra, p. 4 Duality Principle. Let τ and ρ be terms of the augmented language of Boolean algebra. If the universal closure of τ=ρ is a theorem of Boolean algebra, so is the universal closure of τd = ρd. If the universal closure of τ≤ρ is a theorem of Boolean algebra, so is ρd ≤ τd. Proof: Suppose that B is a Boolean algebra and σ is a variable assignment for B that fails to satisfy τd = ρd in B. Let the interpretation A be defined as follows: A = B A(ci) = B(ci), for each i A(“1”) = B(“0”) A(“0”) = B(“1”) x “+”A y = x “⋅”B y x “⋅”A y = x “+”B y “-”A x = “-”B x A(“≤”) = {<y,x>: <x,y> ∈ “≤”B It is easy the verify that, for each term µ and variable assignment δ, d Denδ,A(µ) = Denδ,B(µ ). Consequently, σ fails to satisfy τ=ρ in A. It is also easy to verify, by examining the axioms one by one, that A is a Boolean algebra. So A is a Boolean algebra in which the universal closure of τ=ρ is false. We prove the second part of the Duality Principle — the part about “≤” — by deriving it from the first part. Suppose that the universal closure of τ ≤ ρ is a theorem of Boolean algebra. It follows by the definition of “≤” that the universal closure of (τ+ρ) = ρ is a theorem of Boolean algebra. Using the first part of the Duality Principle, we conclude that the universal closure of (τ+ρ)d = ρd, which is the same as (τd ⋅ ρd) = ρd, is a theorem of Boolean algebra. It follows from the alternative definition of “≤,” together with the commutative law for “⋅,” that ρd ≤ τd is a theorem of Boolean algebra._X We now make the connection between Boolean algebra and the sentential calculus. Let us fix our attention on some SC language whose atomic sentences have been arranged in an infinite list. Definition. For φ an SC sentence whose only connectives are “∨,” “∧,” and “¬,” let us take the algebraic translation of φ, φa, to be the closed term of the augmented language of Boolean algebra, obtained as follows: Boolean algebra, p. 5 Replace the ith atomic sentence, wherever it occurs, by ci, and replace “∨,” “∧,” and “¬,” wherever they occur, by “+,” “⋅,” and “-,” respectively. Definition. If Γ is a set of SC sentences whose only connectives are “∨,” “∧,” and “¬,” Γa = {γa: a ∈ Γ}. Theorem. Let φ and ψ be SC sentences whose only connectives are “∨,” “∧,” and “¬.” We have the following: (a) φ is a tautology iff φa = 1 is a theorem of Boolean algebra. (b) φ is a contradiction iff φa = 0 is a theorem of Boolean algebra. (c) φ and ψ are logically equivalent iff φa = ψa is a theorem of Boolean algebra. (d) φ implies ψ iff φa ≤ ψa is a theorem of Boolean algebra. Proof: We present the proof in the following wacky order: First, we prove the right-to-left direction of (c), then the give the left-to-right direction of (b), and finally we fill in the rest. (c) ⇐ Suppose that φ and ψ are not logically equivalent. Define an interpretation A of the augmented language of Boolean algebra, as follows: For χ an SC sentence, let [χ] be the set of all formulas that are logically equivalent to χ; [χ] is called the logical equivalence class of χ. A = {[χ]: χ an SC formula}. A(ci) = the logical equivalence class of the ith atomic sentence. 1A = the set of tautologies. 0A = the set of inconsistent sentences. For a and b elements of A, define a “+”A b to be the unique member c of A such that there exist sentences χ and θ with a = [χ], b = [θ], and c = [(χ ∨ θ)]. In order for this definition to make sense, we have to persuade ourselves that, for any a and b in A, there is indeed one and only one element c of A that meets the condition; but that’s easy. For a and b elements of A, define a “⋅”A b to be the unique member c of A such that there exist sentences χ and θ with a = [χ], b = [θ], and c = [(χ ∧ θ)]. For a an element of A, define “-”A to be the unique member c of A such that there exists a sentence χ with a = [χ] and c = [¬χ]. It is routine to verify that A, so defined, is a Boolean algebra. For example, the commutative law holds for “+” because, for any sentences χ and θ, (χ ∨ θ) is logically equivalent Boolean algebra, p. 6 to (θ ∨ χ). The associative law holds for “⋅” because, for any sentences χ, θ, and η, ((χ ∧ θ) ∧ η) is logically equivalent to (χ ∧ (θ ∧ η)). And so on. It is also straightforward to verify that, for any formula χ whose only connectives are “∨,” “∧,” and “¬,” A(χa) = [χ]. Since [φ] ≠ [ψ], A is a Boolean algebra in which φa = ψa fails. (b) ⇒ We define a set ∆ of closed terms to be a-inconsistent (for algebraically inconsistent) just in case there exist elements δ1, δ2,..., δn of ∆ such that the sentence (δ1⋅(δ2⋅...⋅δn)...) = 0 is a theorem of Boolean algebra. a-consistent sets will play a role analogous to the role played by d- consistent sets in the proof of the Completeness Theorem. We have the following: Lemma. For any set of closed terms Ω and closed term τ, if Ω ∪ {τ} and Ω ∪ {-τ} are both a-inconsistent, then Ω is a-inconsistent.
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